# 2021-2022 Math Prize Problem #2

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*On the strange island Nagillig, every person has exactly three enemies, no two enemies have a common enemy, and any two non-enemies have exactly one common enemy. How many people are on the island?*

Submit a solution to this problem to Prof. Ken Ching (kching@mmm.edu, CH 614) by **February 23 ^{rd}**. The first correct submission will win a

**first prize of**

**25 dining dollars**. All other correct submissions will receive recognition and bragging rights, and one of these submissions will be randomly selected for a

**second prize of 15 dining dollars**. All are welcome to participate, but prizes are for MMC students only. Solution will be posted online and outside CH 603.

## Solution to 2021-22 Math Prize Problem #2:

There are 10 people on the island. If A is a person on the island, let B, C, and D be the 3 enemies of A. Since no two enemies have a common enemy, no two of the three people, B, C, and D, are enemies. Let B_{1}, B_{2} be the other two enemies of B, and C_{1}, C_{2} the other two enemies of C, and D_{1}, D_{2} the other two enemies of D. These 6 people represent all of the non-enemies of A because any non-enemy of A has exactly one common enemy with A, namely, B, C, or D. Therefore, there are 10 people on the island. To see that the “enemy relationship” holds for B_{1}, B_{2}, C_{1}, C_{2}, D_{1}, D_{2}, we construct the following 6 enemy pairs: B_{1} and C_{1}, B_{1} and D_{1}, B_{2} and C_{2}, B_{2} and D_{2}, C_{1} and D_{2}, C_{2} and D_{1}.

## First Prize Winner: Juliana Tjornhom, Class of `25

Juliana solved the problem by constructing enemy/non-enemy relationships similar to the solution above.